Answer:
A) Radius: 3.44 cm.
Height: 6.88 cm.
B) Radius: 2.73 cm.
Height: 10.92 cm.
Step-by-step explanation:
We have to solve a optimization problem with constraints. The surface area has to be minimized, restrained to a fixed volumen.
a) We can express the volume of the soda can as:

This is the constraint.
The function we want to minimize is the surface, and it can be expressed as:

To solve this, we can express h in function of r:

And replace it in the surface equation

To optimize the function, we derive and equal to zero
![\frac{dS}{dr}=512*(-1)*r^{-2}+4\pi r=0\\\\\frac{-512}{r^2}+4\pi r=0\\\\r^3=\frac{512}{4\pi} \\\\r=\sqrt[3]{\frac{512}{4\pi} } =\sqrt[3]{40.74 }=3.44](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdr%7D%3D512%2A%28-1%29%2Ar%5E%7B-2%7D%2B4%5Cpi%20r%3D0%5C%5C%5C%5C%5Cfrac%7B-512%7D%7Br%5E2%7D%2B4%5Cpi%20r%3D0%5C%5C%5C%5Cr%5E3%3D%5Cfrac%7B512%7D%7B4%5Cpi%7D%20%5C%5C%5C%5Cr%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B512%7D%7B4%5Cpi%7D%20%7D%20%3D%5Csqrt%5B3%5D%7B40.74%20%7D%3D3.44)
The radius that minimizes the surface is r=3.44 cm.
The height is then

The height that minimizes the surface is h=6.88 cm.
b) The new equation for the real surface is:

We derive and equal to zero
![\frac{dS}{dr}=512*(-1)*r^{-2}+8\pi r=0\\\\\frac{-512}{r^2}+8\pi r=0\\\\r^3=\frac{512}{8\pi} \\\\r=\sqrt[3]{\frac{512}{8\pi}}=\sqrt[3]{20.37}=2.73](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdr%7D%3D512%2A%28-1%29%2Ar%5E%7B-2%7D%2B8%5Cpi%20r%3D0%5C%5C%5C%5C%5Cfrac%7B-512%7D%7Br%5E2%7D%2B8%5Cpi%20r%3D0%5C%5C%5C%5Cr%5E3%3D%5Cfrac%7B512%7D%7B8%5Cpi%7D%20%5C%5C%5C%5Cr%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B512%7D%7B8%5Cpi%7D%7D%3D%5Csqrt%5B3%5D%7B20.37%7D%3D2.73)
The radius that minimizes the real surface is r=2.73 cm.
The height is then

The height that minimizes the real surface is h=10.92 cm.
Answer:
yup, it is definitly correct
A, c, d, f
hope this helps you :)
So if we look at the cross section
and assume that the amount turned up is the same on both sides
so what we can do is make 1 equation and find the max or vertex
if the sides are legnth x and base is y
2x+y=18
and the cross sectional area is xy
2x+y=18
minus 2x both sides
y=18-2x
sub for y
x(18-2x)=areamax
18x-2x^2=areamax
we have a parabola
-2x^2+18x+0=areamax
to find the max y valie or area, we find the y value of the vertex
but the x value of the vertex is what we are looking for to find the value of x to max the area
for
ax^2+bx+c=y
x value of vertex is -b/2a
-18/2(-2)=-18/-4=9/2=4.5
4.5=x value of vertex
aswer is 4.5in
C is answer
Every 5 red cars there are 3 Blue cars,
So if there are 40 red cars,
there are going to be 24 Blue cars
40 divided by 5 which is 8
Multiply 3 × 8 which is ( 24 blue cars )